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Find all the eigenvalues and 鈥渆igenvectors鈥 of the linear transformations.

L(A)=A+ATfrom R22toR22 . Isdiagonalizable?

Short Answer

Expert verified

L is diagonalizable and the eigenvalues and eigenvectors of the linear transformation is,

1=0,E2=span1000,01-10,00012=0,E2=span1000,0110,0001

Step by step solution

01

Define eigenvalues

The scalar values that are associated with the vectors of the linear equations in the matrix are called eigenvalues.

Ax=x,here xis eigenvector and is the eigenvalue.

02

Use the required formula and find the eigenvalues and eigenvectors

Consider the equation,

LA=A+AT.

Use the formula,

LA=AAT=-1A

Thus, due to the entries in the main diagonal, this is only possible for 1,2=0,2

Hence, for =0,

role="math" localid="1659596122413" AT=-AA=ab-bdE2=span1000,01-10,0001

Thus, for=2

AT=A

A=abbdE2=span1000,0110,0001

Hence, L is diagonalizable.

Therefore, the eigenvalues and eigenvector is,

1=0,E2=span1000,01-10,00012=0,E2=span1000,0110,0001

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